We discuss the ideas of monotonicity (increasing or decreasing) and concavity (up or down) of a function. Because of the Mean Value Theorem, we can determine these characteristics using derivatives.
Computer Science and Information Science 3rd semester (2012-December) Questio...
Report
Share
1 of 63
Download to read offline
More Related Content
Lesson 21: Derivatives and the Shapes of Curves
1. Section 4.2
Derivatives and the Shapes of Curves
V63.0121.034, Calculus I
November 11, 2009
Announcements
Final Exam Friday, December 18, 2:00–3:50pm
Wednesday, November 25 is a regular class day
.
.
Image credit: cobalt123
. . . . . .
2. Outline
Recall: The Mean Value Theorem
Monotonicity
The Increasing/Decreasing Test
Finding intervals of monotonicity
The First Derivative Test
Concavity
Definitions
Testing for Concavity
The Second Derivative Test
. . . . . .
3. Recall: The Mean Value Theorem
Theorem (The Mean Value
Theorem)
Let f be continuous on [a, b]
and differentiable on (a, b).
Then there exists a point c in
(a, b) such that .
b
.
f(b) − f(a) . .
= f′ (c). a
.
b−a
. . . . . .
4. Recall: The Mean Value Theorem
Theorem (The Mean Value
Theorem)
Let f be continuous on [a, b]
and differentiable on (a, b).
Then there exists a point c in
(a, b) such that .
b
.
f(b) − f(a) . .
= f′ (c). a
.
b−a
. . . . . .
5. Recall: The Mean Value Theorem
Theorem (The Mean Value c
..
Theorem)
Let f be continuous on [a, b]
and differentiable on (a, b).
Then there exists a point c in
(a, b) such that .
b
.
f(b) − f(a) . .
= f′ (c). a
.
b−a
. . . . . .
6. Why the MVT is the MITC
Most Important Theorem In Calculus!
Theorem
Let f′ = 0 on an interval (a, b). Then f is constant on (a, b).
Proof.
Pick any points x and y in (a, b) with x < y. Then f is continuous
on [x, y] and differentiable on (x, y). By MVT there exists a point
z in (x, y) such that
f(y) − f(x)
= f′ (z) = 0.
y−x
So f(y) = f(x). Since this is true for all x and y in (a, b), then f is
constant.
. . . . . .
7. Outline
Recall: The Mean Value Theorem
Monotonicity
The Increasing/Decreasing Test
Finding intervals of monotonicity
The First Derivative Test
Concavity
Definitions
Testing for Concavity
The Second Derivative Test
. . . . . .
9. What does it mean for a function to be increasing?
Definition
A function f is increasing on (a, b) if
f(x) < f(y)
whenever x and y are two points in (a, b) with x < y.
An increasing function “preserves order.”
Write your own definition (mutatis mutandis) of decreasing,
nonincreasing, nondecreasing
. . . . . .
10. The Increasing/Decreasing Test
Theorem (The Increasing/Decreasing Test)
If f′ > 0 on (a, b), then f is increasing on (a, b). If f′ < 0 on (a, b),
then f is decreasing on (a, b).
. . . . . .
11. The Increasing/Decreasing Test
Theorem (The Increasing/Decreasing Test)
If f′ > 0 on (a, b), then f is increasing on (a, b). If f′ < 0 on (a, b),
then f is decreasing on (a, b).
Proof.
It works the same as the last theorem. Pick two points x and y in
(a, b) with x < y. We must show f(x) < f(y). By MVT there exists
a point c in (x, y) such that
f(y) − f(x)
= f′ (c) > 0.
y−x
So
f(y) − f(x) = f′ (c)(y − x) > 0.
. . . . . .
13. Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f(x) = 2x − 5.
Solution
f′ (x) = 2 is always positive, so f is increasing on (−∞, ∞).
. . . . . .
14. Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f(x) = 2x − 5.
Solution
f′ (x) = 2 is always positive, so f is increasing on (−∞, ∞).
Example
Describe the monotonicity of f(x) = arctan(x).
. . . . . .
15. Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f(x) = 2x − 5.
Solution
f′ (x) = 2 is always positive, so f is increasing on (−∞, ∞).
Example
Describe the monotonicity of f(x) = arctan(x).
Solution
1
Since f′ (x) = is always positive, f(x) is always increasing.
1 + x2
. . . . . .
17. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
f′ (x) = 2x, which is positive when x > 0 and negative when x
is.
. . . . . .
18. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
f′ (x) = 2x, which is positive when x > 0 and negative when x
is.
We can draw a number line:
−
. 0
.. .
+ .′
f
0
.
. . . . . .
19. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
f′ (x) = 2x, which is positive when x > 0 and negative when x
is.
We can draw a number line:
−
. 0
.. .
+ .′
f
↘
. 0
. ↗
. f
.
. . . . . .
20. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
f′ (x) = 2x, which is positive when x > 0 and negative when x
is.
We can draw a number line:
−
. 0
.. .
+ .′
f
↘
. 0
. ↗
. f
.
So f is decreasing on (−∞, 0) and increasing on (0, ∞).
. . . . . .
21. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
f′ (x) = 2x, which is positive when x > 0 and negative when x
is.
We can draw a number line:
−
. 0
.. .
+ .′
f
↘
. 0
. ↗
. f
.
So f is decreasing on (−∞, 0) and increasing on (0, ∞).
In fact we can say f is decreasing on (−∞, 0] and increasing
on [0, ∞)
. . . . . .
23. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solution
f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 1 x−1/3 (5x + 4)
3 3
The critical points are 0 and and −4/5.
−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
.x+4
5
−
. 4/5
. . . . . .
24. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solution
f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 1 x−1/3 (5x + 4)
3 3
The critical points are 0 and and −4/5.
−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
.x+4
5
−
. 4/5
0
.. ×
.. .′ (x)
f
−
. 4/5 0
. f
.(x)
. . . . . .
25. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solution
f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 1 x−1/3 (5x + 4)
3 3
The critical points are 0 and and −4/5.
−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
.x+4
5
−
. 4/5
.
+ 0 − ×
.. . . . .
+ .′ (x)
f
↗
. − ↘ .
. 4/5 . 0 ↗
. f
.(x)
. . . . . .
26. The First Derivative Test
Theorem (The First Derivative Test)
Let f be continuous on [a, b] and c a critical point of f in (a, b).
If f′ (x) > 0 on (a, c) and f′ (x) < 0 on (c, b), then c is a local
maximum.
If f′ (x) < 0 on (a, c) and f′ (x) > 0 on (c, b), then c is a local
minimum.
If f′ (x) has the same sign on (a, c) and (c, b), then c is not a
local extremum.
. . . . . .
27. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
f′ (x) = 2x, which is positive when x > 0 and negative when x
is.
We can draw a number line:
−
. 0
.. .
+ .′
f
↘
. 0
. ↗
. f
.
So f is decreasing on (−∞, 0) and increasing on (0, ∞).
In fact we can say f is decreasing on (−∞, 0] and increasing
on [0, ∞)
. . . . . .
28. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
f′ (x) = 2x, which is positive when x > 0 and negative when x
is.
We can draw a number line:
−
. 0
.. .
+ .′
f
↘
. 0
. ↗
. f
.
m
. in
So f is decreasing on (−∞, 0) and increasing on (0, ∞).
In fact we can say f is decreasing on (−∞, 0] and increasing
on [0, ∞)
. . . . . .
29. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solution
f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 1 x−1/3 (5x + 4)
3 3
The critical points are 0 and and −4/5.
−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
.x+4
5
−
. 4/5
.
+ 0 − ×
.. . . . .
+ .′ (x)
f
↗
. − ↘ .
. 4/5 . 0 ↗
. f
.(x)
. . . . . .
30. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solution
f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 1 x−1/3 (5x + 4)
3 3
The critical points are 0 and and −4/5.
−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
.x+4
5
−
. 4/5
.
+ 0 − ×
.. . . . .
+ .′ (x)
f
↗
. − ↘ .
. 4/5 . 0 ↗
. f
.(x)
m
. ax . in
m
. . . . . .
31. Outline
Recall: The Mean Value Theorem
Monotonicity
The Increasing/Decreasing Test
Finding intervals of monotonicity
The First Derivative Test
Concavity
Definitions
Testing for Concavity
The Second Derivative Test
. . . . . .
32. Definition
The graph of f is called concave up on and interval I if it lies
above all its tangents on I. The graph of f is called concave down
on I if it lies below all its tangents on I.
. .
concave up concave down
We sometimes say a concave up graph “holds water” and a
concave down graph “spills water”.
. . . . . .
33. Definition
A point P on a curve y = f(x) is called an inflection point if f is
continuous there and the curve changes from concave upward to
concave downward at P (or vice versa).
.concave up
i
.nflection point
. .
.
concave
down
. . . . . .
34. Theorem (Concavity Test)
If f′′ (x) > 0 for all x in an interval I, then the graph of f is
concave upward on I
If f′′ (x) < 0 for all x in I, then the graph of f is concave
downward on I
. . . . . .
35. Theorem (Concavity Test)
If f′′ (x) > 0 for all x in an interval I, then the graph of f is
concave upward on I
If f′′ (x) < 0 for all x in I, then the graph of f is concave
downward on I
Proof.
Suppose f′′ (x) > 0 on I. This means f′ is increasing on I. Let a and
x be in I. The tangent line through (a, f(a)) is the graph of
L(x) = f(a) + f′ (a)(x − a)
f(x) − f(a)
By MVT, there exists a c between a and x with = f′ (c).
x−a
So
f(x) = f(a) + f′ (c)(x − a) ≥ f(a) + f′ (a)(x − a) = L(x)
. . . . . .
39. Example
Find the intervals of concavity for the graph of f(x) = x3 + x2 .
Solution
We have f′ (x) = 3x2 + 2x, so f′′ (x) = 6x + 2.
This is negative when x < −1/3, positive when x > −1/3, and
0 when x = −1/3
So f is concave down on (−∞, −1/3), concave up on
(1/3, ∞), and has an inflection point at (−1/3, 2/27)
. . . . . .
43. Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
Solution
10 −1/3 4 −4/3 2 −4/3
f′′ (x) = x − x = x (5x − 2)
9 9 9
The second derivative f′′ (x) is not defined at 0
Otherwise, x−4/3 is always positive, so the concavity is
determined by the 5x − 2 factor
. . . . . .
44. Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
Solution
10 −1/3 4 −4/3 2 −4/3
f′′ (x) = x − x = x (5x − 2)
9 9 9
The second derivative f′′ (x) is not defined at 0
Otherwise, x−4/3 is always positive, so the concavity is
determined by the 5x − 2 factor
So f is concave down on (−∞, 0], concave down on [0, 2/5),
concave up on (2/5, ∞), and has an inflection point when
x = 2/5
. . . . . .
45. The Second Derivative Test
Theorem (The Second Derivative Test)
Let f, f′ , and f′′ be continuous on [a, b]. Let c be be a point in
(a, b) with f′ (c) = 0.
If f′′ (c) < 0, then f(c) is a local maximum.
If f′′ (c) > 0, then f(c) is a local minimum.
. . . . . .
46. The Second Derivative Test
Theorem (The Second Derivative Test)
Let f, f′ , and f′′ be continuous on [a, b]. Let c be be a point in
(a, b) with f′ (c) = 0.
If f′′ (c) < 0, then f(c) is a local maximum.
If f′′ (c) > 0, then f(c) is a local minimum.
Remarks
If f′′ (c) = 0, the second derivative test is inconclusive (this
does not mean c is neither; we just don’t know yet).
We look for zeroes of f′ and plug them into f′′ to determine if
their f values are local extreme values.
. . . . . .
50. Example
Find the local extrema of f(x) = x3 + x2 .
Solution
f′ (x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.
Remember f′′ (x) = 6x + 2
Since f′′ (−2/3) = −2 < 0, −2/3 is a local maximum.
. . . . . .
51. Example
Find the local extrema of f(x) = x3 + x2 .
Solution
f′ (x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.
Remember f′′ (x) = 6x + 2
Since f′′ (−2/3) = −2 < 0, −2/3 is a local maximum.
Since f′′ (0) = 2 > 0, 0 is a local minimum.
. . . . . .
54. Example
Find the local extrema of f(x) = x2/3 (x + 2)
Solution
1 −1/3
Remember f′ (x) = x (5x + 4) which is zero when
3
x = −4/5
10 −4/3
Remember f′′ (x) = x (5x − 2), which is negative when
9
x = −4/5
. . . . . .
55. Example
Find the local extrema of f(x) = x2/3 (x + 2)
Solution
1 −1/3
Remember f′ (x) = x (5x + 4) which is zero when
3
x = −4/5
10 −4/3
Remember f′′ (x) = x (5x − 2), which is negative when
9
x = −4/5
So x = −4/5 is a local maximum.
. . . . . .
56. Example
Find the local extrema of f(x) = x2/3 (x + 2)
Solution
1 −1/3
Remember f′ (x) = x (5x + 4) which is zero when
3
x = −4/5
10 −4/3
Remember f′′ (x) = x (5x − 2), which is negative when
9
x = −4/5
So x = −4/5 is a local maximum.
Notice the Second Derivative Test doesn’t catch the local
minimum x = 0 since f is not differentiable there.
. . . . . .
58. When the second derivative is zero
At inflection points c, if f′ is differentiable at c, then f′′ (c) = 0
Is it necessarily true, though?
. . . . . .
59. When the second derivative is zero
At inflection points c, if f′ is differentiable at c, then f′′ (c) = 0
Is it necessarily true, though?
Consider these examples:
f (x ) = x 4 g(x) = −x4 h(x) = x3
. . . . . .
60. When first and second derivative are zero
function derivatives graph type
f′ (x) = 4x3 , f′ (0) = 0
f (x ) = x 4 min
f′′ (x) = 12x2 , f′′ (0) = 0 .
.
g′ (x) = −4x3 , g′ (0) = 0
g (x ) = −x4 max
g′′ (x) = −12x2 , g′′ (0) = 0
h′ (x) = 3x2 , h′ (0) = 0
h(x) = x3 infl.
h′′ (x) = 6x, h′′ (0) = 0 .
. . . . . .
61. When the second derivative is zero
At inflection points c, if f′ is differentiable at c, then f′′ (c) = 0
Is it necessarily true, though?
Consider these examples:
f (x ) = x 4 g(x) = −x4 h(x) = x3
All of them have critical points at zero with a second derivative of
zero. But the first has a local min at 0, the second has a local
max at 0, and the third has an inflection point at 0. This is why
we say 2DT has nothing to say when f′′ (c) = 0.
. . . . . .
62. What have we learned today?
Concepts: Mean Value Theorem, monotonicity, concavity
Facts: derivatives can detect monotonicity and concavity
Techniques for drawing curves: the Increasing/Decreasing
Test and the Concavity Test
Techniques for finding extrema: the First Derivative Test and
the Second Derivative Test
. . . . . .
63. What have we learned today?
Concepts: Mean Value Theorem, monotonicity, concavity
Facts: derivatives can detect monotonicity and concavity
Techniques for drawing curves: the Increasing/Decreasing
Test and the Concavity Test
Techniques for finding extrema: the First Derivative Test and
the Second Derivative Test
Next week: Graphing functions!
. . . . . .